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Difference between revisions of "2021 Fall AMC 10B Problems/Problem 8"

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==Solution 2==
 
==Solution 2==
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Since <math>16384</math> is <math>2^14</math>, we can consider it as <math>(2^7)^2</math>. <math>16383</math> is <math>1</math> less than <math>16384</math>, so it can be considered as <math>1</math> less than a square. Therefore, it can be expressed as <math>(x-1)(x+1)</math>. Since <math>2^7</math> is <math>128, 16383</math> is <math>127 \cdot 129</math>. <math>129</math> is <math>3 \cdot 43</math>, and since <math>127</math> is larger, our answer is <math>\boxed {(C) 10}</math>.
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~Arcticturn
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021 Fall|ab=B|num-a=9|num-b=7}}
 
{{AMC10 box|year=2021 Fall|ab=B|num-a=9|num-b=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:38, 23 November 2021

Problem 8

The largest prime factor of $16384$ is $2$, because $16384 = 2^{14}$. What is the sum of the digits of the largest prime factor of $16383$?

$\textbf{(A) }3\qquad\textbf{(B) }7\qquad\textbf{(C) }10\qquad\textbf{(D) }16\qquad\textbf{(E) }22$

Solution 1

We have

\[16383=16384-1=2^{14}-1=(2^7+1)(2^7-1)=129\cdot127=3\cdot43\cdot127.\] Since $127$ is prime, our answer is $\boxed{\textbf{(C) }10}$.

~kingofpineapplz

Solution 2

Since $16384$ is $2^14$, we can consider it as $(2^7)^2$. $16383$ is $1$ less than $16384$, so it can be considered as $1$ less than a square. Therefore, it can be expressed as $(x-1)(x+1)$. Since $2^7$ is $128, 16383$ is $127 \cdot 129$. $129$ is $3 \cdot 43$, and since $127$ is larger, our answer is $\boxed {(C) 10}$.

~Arcticturn

See Also

2021 Fall AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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